Problem 87
Question
Let \(g(x)=\int_{0}^{x} f(t) d t\), where \(f\) is such that \(\frac{1}{2} \leq f(t) \leq 1\) for \(t \in[0,1]\) and \(0 \leq f(t) \leq \frac{1}{2}\) for \(t \in[1,2]\). Then, (A) \(-\frac{3}{2} \leq g(2) \leq \frac{1}{2}\) (B) \(\frac{3}{2} \leq g(2) \leq \frac{5}{2}\) (C) \(\frac{1}{2} \leq g(2) \leq \frac{3}{2}\) (D) None of these
Step-by-Step Solution
Verified Answer
The answer is (C).
1Step 1: Analyze the function on interval [0,1]
For \(t \in [0, 1]\), the function \(f(t)\) is bounded by \(\frac{1}{2} \leq f(t) \leq 1\). This means that the integral \(\int_{0}^{1} f(t) \, dt\) can be estimated as follows:\[\int_{0}^{1} \frac{1}{2} \, dt \leq \int_{0}^{1} f(t) \, dt \leq \int_{0}^{1} 1 \, dt\]Calculating these bounds:- Lower bound: \(\int_{0}^{1} \frac{1}{2} \, dt = \frac{1}{2} \cdot (1 - 0) = \frac{1}{2}\)- Upper bound: \(\int_{0}^{1} 1 \, dt = 1 \cdot (1 - 0) = 1\)
2Step 2: Analyze the function on interval [1,2]
For \(t \in [1, 2]\), the function \(f(t)\) is bounded by \(0 \leq f(t) \leq \frac{1}{2}\). This leads to the following estimate for the integral \(\int_{1}^{2} f(t) \, dt\):\[\int_{1}^{2} 0 \, dt \leq \int_{1}^{2} f(t) \, dt \leq \int_{1}^{2} \frac{1}{2} \, dt\]Calculating these bounds:- Lower bound: \(\int_{1}^{2} 0 \, dt = 0 \cdot (2 - 1) = 0\)- Upper bound: \(\int_{1}^{2} \frac{1}{2} \, dt = \frac{1}{2} \cdot (2 - 1) = \frac{1}{2}\)
3Step 3: Compute overall bounds for g(2)
Now, combine the results from Step 1 and Step 2 to find bounds for \(g(2) = \int_{0}^{2} f(t) \, dt\):\[\int_{0}^{1} f(t) \, dt + \int_{1}^{2} f(t) \, dt = g(2)\]Using the bounds:- Lower bound: \(\frac{1}{2} + 0 = \frac{1}{2}\)- Upper bound: \(1 + \frac{1}{2} = \frac{3}{2}\)
4Step 4: Compare to given options
The computed range for \(g(2)\) is \(\frac{1}{2} \leq g(2) \leq \frac{3}{2}\). Compare this with the options:- Option (A): \(-\frac{3}{2} \leq g(2) \leq \frac{1}{2}\) is not satisfied, as the upper bound is too low.- Option (B): \(\frac{3}{2} \leq g(2) \leq \frac{5}{2}\) is not satisfied, as the lower bound is too high.- Option (C): \(\frac{1}{2} \leq g(2) \leq \frac{3}{2}\) matches the computed bounds exactly.Therefore, the correct answer is (C).
Key Concepts
Fundamental Theorem of CalculusDefinite IntegralsFunction Analysis
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a cornerstone in the understanding of integral calculus. It connects differentiation and integration, two major concepts in calculus. This theorem can be understood in two parts.
First, it states that if you have a continuous function, the integral of its derivative over an interval gives you the net change in the function over that interval. This means if you differentiate a function and then integrate that derivative, you end up back where you started. In simpler terms, doing an integral of a derivative retrieves the original function, up to a constant.
Second, the theorem asserts that if you calculate the integral of a function over an interval, it gives you a function whose derivative is the original function. Essentially, integrating involves computing an antiderivative. This helps in finding the definite integral, or total accumulation, of a function over an interval.
First, it states that if you have a continuous function, the integral of its derivative over an interval gives you the net change in the function over that interval. This means if you differentiate a function and then integrate that derivative, you end up back where you started. In simpler terms, doing an integral of a derivative retrieves the original function, up to a constant.
Second, the theorem asserts that if you calculate the integral of a function over an interval, it gives you a function whose derivative is the original function. Essentially, integrating involves computing an antiderivative. This helps in finding the definite integral, or total accumulation, of a function over an interval.
Definite Integrals
Definite integrals are used to calculate the total accumulation of a quantity, often represented as the area under a curve. If you imagine sketching a graph of a function over a specific interval, the definite integral calculates the area between this curve and the horizontal axis within that interval.
The notation for a definite integral, such as \ \( \int_{a}^{b} f(t) \, dt \ \), consists of:
The notation for a definite integral, such as \ \( \int_{a}^{b} f(t) \, dt \ \), consists of:
- \(f(t)\), the integrand, which is the function you're integrating.
- The limits of integration \([a, b]\), indicating the interval over which you're integrating.
- \(dt\) shows the variable with respect to which you're integrating.
Function Analysis
Function analysis involves understanding the behavior and properties of functions. This includes exploring how functions behave over intervals and how their values are bounded. For function analysis in calculus, you often determine how a function's graph behaves and what this behavior says about integral limits.
In the given exercise, the analysis began by observing the function \(f(t)\) within specific intervals, \([0, 1]\) and \([1, 2]\). For each interval, we determined how \(f(t)\) is bounded, meaning the minimum and maximum values it can take. This kind of analysis helps to set the bounds for integrals, as seen during the computations for \(g(2)\), where the integral was split over the balanced intervals.
A key point in function analysis involves calculating the bounds of the definite integrals separately within each interval and then combining them.
This aids in understanding the entire behavior of the function over the larger interval.
In the given exercise, the analysis began by observing the function \(f(t)\) within specific intervals, \([0, 1]\) and \([1, 2]\). For each interval, we determined how \(f(t)\) is bounded, meaning the minimum and maximum values it can take. This kind of analysis helps to set the bounds for integrals, as seen during the computations for \(g(2)\), where the integral was split over the balanced intervals.
A key point in function analysis involves calculating the bounds of the definite integrals separately within each interval and then combining them.
This aids in understanding the entire behavior of the function over the larger interval.
Other exercises in this chapter
Problem 85
\(f(x)\) is a continuous function for all real values of \(x\) and satisfies \(\int_{0}^{x} f(t) d t=\int_{x}^{1} t^{2} f(t) d t+\frac{x^{16}}{8}+\frac{x^{6}}{3
View solution Problem 86
If \(f(x)=\frac{x-1}{x+1}, f^{2}(x)=f(f(x)), \ldots f_{(x)}^{k+1}=f\left(f^{k}(x)\right)\), \(k=1,2,3, \ldots\) and \(\phi(x)=f^{1998}(x)\), then \(\int_{1 / e}
View solution Problem 88
\(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sin ^{2 k} \frac{r \pi}{2 n}\) is equal to (A) \(\frac{2 k !}{2^{2 k}(k !)^{2}}\) (B) \(\frac{2 k !}{
View solution Problem 90
The value of the integral \(\int_{0}^{41 \pi / 4}|\cos x| d x\) is (A) \(20-\frac{1}{\sqrt{2}}\) (B) \(20+\frac{1}{\sqrt{2}}\) (C) \(19+\frac{1}{\sqrt{2}}\) (D)
View solution